已知sin(π/4-a)=5/13(0<a<π/4),求cos(π/4+a)/cos2a
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已知sin(π/4-a)=5/13(0<a<π/4),求cos(π/4+a)/cos2a
已知sin(π/4-a)=5/13(0<a<π/4),求cos(π/4+a)/cos2a
已知sin(π/4-a)=5/13(0<a<π/4),求cos(π/4+a)/cos2a
sin(π/4-a)=5/13=cos(π/4+a)
0<2a<π/2
sin2a=cos(π/2-2a)=1-2(sin(π/4-a))^2=1-50/169=119/169
cos2a=√(288*50)/169≡120/169
(5/13)/(120/169)=13/24
cos(π/4+a)/cos2a=cos(π/4+a)/sin[2(a+π/4)]=1/2sin(π/4+a)=1/2cos(π/4-a)。
因sin(π/4-a)=5/13,则cos(π/4-a)=12/13,即这个式子的值就是13/24